按名称过滤列后填充nan的数据

Question

有谁能告诉我在这里根据另一个列名过滤(和填充nan)的正确方法。非常感谢。

相关链接：How to fill dataframe's empty/nan cell with conditional column mean

df

ID  Name        Industry            Expenses
1   Treslam     Financial Services  734545
2   Rednimdox   Construction        nan
3   Lamtone     IT Services         567678
4   Stripfind   Financial Services  nan
5   Openjocon   Construction        8678957
6   Villadox    Construction        5675676
7   Sumzoomit   Construction        231244
8   Abcd        Construction        nan
9   Stripfind   Financial Services  nan


df_mean_expenses = (df.groupby(['Industry'], as_index = False)['Expenses']).mean()
df_mean_expenses

  Industry           Expenses
0 Construction       554433.11
1 Financial Services 2362818.48
2 IT Services        149153.46

为了用结构行的平均值(在df_mean_expenses中)替换结构-收入nan，我尝试了两种方法：

1. df.loc[df'Expenses'.isna()，‘费用’[df‘’Industry‘==’Construction‘= df_mean_expenses.loc[df_mean_expenses'Industry’== 'Construction'，'Expenses'].values

。。返回错误:项目错误长度500，而不是3！

2.

df'Expenses'np.isnan(df'Expenses')][df'Industry‘== 'Construction'] = df_mean_expenses.loc[df_mean_expenses'Industry’== 'Construction'，‘Construction’.values .这将运行，但不会向df添加值。

预期产出：

df

 ID Name        Industry            Expenses
    1   Treslam     Financial Services  734545
    2   Rednimdox   Construction        554433.11
    3   Lamtone     IT Services         567678
    4   Stripfind   Financial Services  nan
    5   Openjocon   Construction        8678957
    6   Villadox    Construction        5675676
    7   Sumzoomit   Construction        231244
    8   Abcd        Construction        554433.11
    9   Stripfind   Financial Services  nan

Answer 1

尝试使用transform

df_mean_expenses = df.groupby('Industry')['Expenses'].transform('mean')
df['Revenue'] = df['Revenue'].fillna(df_mean_expenses[df['Industry']=='Construction'])